Answer
The function $u=1/\sqrt{x^2+y^2+z^2}$ is a solution of the equation $u_{xx}+u_{yy}+u_{zz}=0$.
Work Step by Step
Here, we have $u_x=-\dfrac{x}{(x^2+y^2+z^2)^{3/2}}\\u_{xx}=\dfrac{2x^2-y^2-z^2}{(x^2+y^2+z^2)^{5/2}}$ ...(A)
and $u_{yy}=\dfrac{(2y^2-x^2-z^2)}{(x^2+y^2+z^2)^{5/2}}$ ...(B)
Also, $u_{zz}=\dfrac{(2z^2-y^2-x^2)}{(x^2+y^2+z^2)^{5/2}}$ ...(C)
We can see from the equations (A), (B) and (C) that
$u_{xx}+u_{yy}+u_{zz}=\dfrac{(2x^2-y^2-z^2)+(2y^2-x^2-z^2)+(2z^2-y^2-x^2)}{(x^2+y^2+z^2)^{5/2}}=0$
Thus, it has been verified that the function $u=1/\sqrt{x^2+y^2+z^2}$ is a solution of the equation $u_{xx}+u_{yy}+u_{zz}=0$.